Answer:
20kg of 25% copper alloy
30kg of 60% copper alloy
Step-by-step explanation:
There are 2 kinds of metal, first metal(A) have 25% copper while the second metal(B) has 60% copper. The metalworker want to create 50kg of metal, which means the total weight of both metals is 50kg (A+B = 50). The metalworker also want the metal made of a 46% which is 23kg(0.25A + 0.6B = 0.46*50). From these sentences, we can derive 2 equations. We can solve this with substitution.
A+B = 50
A= 50-B
Let's put the first equation into the second.
0.25A + 0.6B = 0.46 *50
0.25(50-B) + 0.6B =23
12.5 - 0.25B + 0.6B =23
0.35B=23 -12.5
B= 10.5/ 0.35= 30
Then we can solve A
A= 50-B
A= 50-30
A=20
Answer:
8x³ + 12x² - 16x - 16
General Formulas and Concepts:
- Order of Operations: BPEMDAS
- Distributive Property
Step-by-step explanation:
<u>Step 1: Define expression</u>
(4x² - 2x - 4)(2x + 4)
<u>Step 2: Simplify</u>
- Expand: 8x³ - 4x² - 8x + 16x² - 8x - 16
- Combine like terms (x²): 8x³ + 12x² - 8x - 8x - 16
- Combine like terms (x): 8x³ + 12x² - 16x - 16
Answer:
1 1/24
Step-by-step explanation:
5/12 + 5/8
We need a common denominator of 24
Convert the first number to have a denominator of 24
5/12 *2/2 = 10/24
Convert the second number to have a denominator of 24
5/8*3/3 = 15/24
Add the two converted numbers together
10/24 + 15/24
25/24
Change from an improper fraction to a mixed number
24/24 + 1/24
1 1/24
Answer:
129/2
Step-by-step explanation:
1 129
64 --- = -------
2 2
To find the improper fraction, you multiply the denominator by the whole number and then add the numerator. In the end, you put that answer over the starting denominator.
64 X 2 = 128 128 + 1 = 129 129 / 2 is the answer
Here is one <span>Jake’s salary depends on the number of hours he works.
The independent variable is the number of hours and the dependent variable is salary.
Let x = the number of hours worked
Let y = Jake's salary
The set of ordered pairs {(1, 10), (2, 20), (3, 30), (4, 40), (5, 50)} can be used to represent
the function, assuming Jake earns $10 per hour.
</span>
